Question: Simplify and expand the following expression: $ \dfrac{5z + 2}{4z - 9}-\dfrac{3z}{z + 7} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(4z - 9)(z + 7)$ Multiply the first term by $\dfrac{z + 7}{z + 7}$ $ \begin{align*} \dfrac{5z + 2}{4z - 9} \times \dfrac{z + 7}{z + 7} & = \dfrac{(5z + 2)(z + 7)}{(4z - 9)(z + 7)} \\ & = \dfrac{5z^2 + 37z + 14}{(4z - 9)(z + 7)}\end{align*} $ Multiply the second term by $\dfrac{4z - 9}{4z - 9}$ $ \begin{align*} \dfrac{3z}{z + 7} \times \dfrac{4z - 9}{4z - 9} & = \dfrac{(3z)(4z - 9)}{(z + 7)(4z - 9)} \\ & = \dfrac{12z^2 - 27z}{(z + 7)(4z - 9)}\end{align*} $ Now we have: $ = \dfrac{5z^2 + 37z + 14}{(4z - 9)(z + 7)} - \dfrac{12z^2 - 27z}{(z + 7)(4z - 9)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5z^2 + 37z + 14 - (12z^2 - 27z)}{(4z - 9)(z + 7)} $ $ = \dfrac{5z^2 + 37z + 14 - 12z^2 + 27z}{(4z - 9)(z + 7)} $ $ = \dfrac{-7z^2 + 64z + 14}{(4z - 9)(z + 7)}$ Expand the denominator: $ = \dfrac{-7z^2 + 64z + 14}{4z^2 + 19z - 63}$